Storing command values into a variable Bash Script - linux

Question

I am trying to loop through files in a directory to find an animal and its value. The command is supposed to only display the animal and total value. For example:
File1 has:
Monkey 11
Bear 4
File2 has:
Monkey 12
If I wanted the total value of monkeys then I would do:
for f in *; do
total=$(grep $animal $f | cut -d " " -f 2- | paste -sd+ | bc)
done
echo $animal $total
This would return the correct value of:
Monkey 23
However, if there is only one instance of an animal like for example Bear, the variable total doesn't return any value, I only get echoed:
Bear
Why is this the case and how do I fix it?
Note: I'm not allowed to use the find command.

Answer 1

you could use this little awk instead of for grep cut paste bc:
awk -v animal="Bear" '
$1 == animal { count += $2 }
END { print count + 0 }
' *

Answer 2

Comments on OP's question about why code behaves as it does:
total is reset on each pass through the loop so ...
upon leaving the loop total will have the count from the 'last' file processed
in the case of Bear the 'last' file processed is File2 and since File2 does not contain any entries for Bear we get total='', which is what's printed by the echo
if the Bear entry is moved from File1 to File2 then OP's code should print Bear 4
OP's current code effectively ignores all input files and prints whatever's in the 'last' file (File2 in this case)
OP's current code generates the following:
# Monkey
Monkey 12 # from File2
# Bear
Bear # no match in File2
I'd probably opt for replacing the whole grep/cut/paste/bc (4x subprocesses) with a single awk (1x subprocess) call (and assuming no matches we report 0):
for animal in Monkey Bear Hippo
do
total=$(awk -v a="${animal}" '$1==a {sum+=$2} END {print sum+0}' *)
echo "${animal} ${total}"
done
This generates:
Monkey 23
Bear 4
Hippo 0
NOTES:
I'm assuming OP's real code does more than echo the count to stdout hence the need of the total variable otherwise we could eliminate the total variable and have awk print the animal/sum pair directly to stdout
if OP's real code has a parent loop processing a list of animals it's likely possible a single awk call could process all of the animals at once; objective being to have awk generate the entire set of animal/sum pairs that could then be fed to the looping construct; if this is the case, and OP has some issues implementing a single awk solution, a new question should be asked

Answer 3

Why is this the case
grep outputs nothing, so nothing is propagated through the pipe and empty string is assigned to total.
Because total is reset every loop (total=anything without referencing previous value), it just has the value for the last file.
how do I fix it?
Do not try to do all at once, just less thing at once.
total=0
for f in *; do
count=$(grep "$animal" "$f" | cut -d " " -f 2-)
total=$((total + count)) # reuse total, reference previous value
done
echo "$animal" "$total"
A programmer fluent in shell will most probably jump to AWK for such problems. Remember to check your scripts with shellcheck.
With what you were trying to do, you could do all files at once:
total=$(
{
echo 0 # to have at least nice 0 if animal is not found
grep "$animal" * |
cut -d " " -f 2-
} |
paste -sd+ |
bc
)

Answer 4

With just bash:
declare -A animals=()
for f in *; do
while read -r animal value; do
(( animals[$animal] = ${animals[$animal]:-0} + value ))
done < "$f"
done
declare -p animals
outputs
declare -A animals=([Monkey]="23" [Bear]="4" )
With this approach, you have all the totals for all the animals by processing each file exactly once

Answer 5

$ head File*
==> File1 <==
Monkey 11
Bear 4
==> File2 <==
Monkey 12
==> File3 <==
Bear
Monkey
Using awk and bash array
#!/bin/bash
sumAnimals(){
awk '
{ NF == 1 ? a[$1]++ : a[$1]=a[$1]+$2 }
END{
for (i in a ) printf "[%s]=%d\n",i, a[i]
}
' File*
}
# storing all animals in bash array
declare -A animalsArr="( $(sumAnimals) )"
# show array content
declare -p animalsArr
# getting total from array
echo "Monkey: ${animalsArr[Monkey]}"
echo "Bear: ${animalsArr[Monkey]}"
Output
declare -A animalsArr=([Bear]="5" [Monkey]="24" )
Monkey: 24
Bear: 5